Alexandr Belov Alexandr Belov - 2 months ago 40
HTML Question

Unrecognized expression: a[href^=#]

I am trying to make a trivial "scrollTo element" function.

Console shows me the error:
Syntax error, unrecognized expression: a[href^=#] for a[href^=#] code

According to the reply on this question I wrapped the hash symbol into double quotes, but now console shows Unexpected token ILLEGAL for that.

Please, explain what I am doing wrong and how to fix it.

Here's my code:

$(document).on('click', 'a[href^=#]', function () {
$('html, body').animate({ scrollTop: $('section[data-target="'+this.hash.slice(1)+'"]').offset().top }, 1000 );
return false;

menu {
background-color: #1abc9c;
height: 50px;
position: fixed;
left: 0;
top: 0;
width: 100%;
margin: 0;
padding: 0;

menu ul li {
display: inline-block;
padding: 0 15px;

menu ul li a {
color: #333;
text-decoration: none;

section {
height: 300px;
padding: 60px 0 0 45px;

.one {
background-color: #3498db;

.two {
background-color: #e74c3c;

.three {
background-color: #f39c12;

.four {
background-color: #2c3e50;

<li><a href="#one">One</a></li>
<li><a href="#two">Two</a></li>
<li><a href="#three">Three</a></li>
<li><a href="#four">Four</a></li>

<section class="one" data-target="one">Section One</section>
<section class="two" data-target="two">Section Two</section>
<section class="three" data-target="three">Section Three</section>
<section class="four" data-target="four">Section Four</section>

and the same in JSFiddle

Answer Source

Since # is a metacharacter of jquery, you have to wrap it by using quotes or you have to escape it by using \\,

$(document).on('click', 'a[href^="#"]', function () {


$(document).on('click', 'a[href^=\\#]', function () {